Given that,
Mass of a tribble, m = 2.5 kg
Radius, r = 1.4 m
The force on the tribble from the bucket does not exceed 10 times its weight.
To find,
The maximum tangential speed.
Solution,
The force acting on the tribble is equal to the centripetal force.
F = 10mg
The formula for the centripetal force is given by :
v is maximum tangential speed
So, the maximum tangential speed is 11.7 m/s.
Answer:
It can be concluded that the star is moving away from the observer.
Explanation:
Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).
The wavelength at rest for this case is 434 nm and 410 nm (,
)
Since, (444nm) is greater than
(434 nm) and
(420nm) is greater than
(410 nm), it can be concluded that the star is moving away from the observer
Answer:
The temperature of the steam during the heat rejection process is 42.5°C
Explanation:
Given the data in the question;
the maximum temperature T in the cycle is twice the minimum absolute temperature T
in the cycle
T = 0.5T
now, we find the efficiency of the Carnot cycle engine
η = 1 - T
/T
η = 1 - T
/0.5T
η = 0.5
the efficiency of the Carnot heat engine can be expressed as;
η = 1 - W
/Q
where W is net work done, Q
is is the heat supplied
we substitute
0.5 = 60 / Q
Q = 60 / 0.5
Q = 120 kJ
Now, we apply the first law of thermodynamics to the system
W = Q
- Q
60 = 120 - Q
Q = 60 kJ
now, the amount of heat rejection per kg of steam is;
q = Q
/m
we substitute
q = 60/0.025
q = 2400 kJ/kg
which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)
q = h
= 2400 kJ/kg
now, at h = 2400 kJ/kg from saturated water tables;
T = 40 + ( 45 - 40 ) (
)
T = 40 + (5) × (0.5)
T = 40 + 2.5
T = 42.5°C
Therefore, The temperature of the steam during the heat rejection process is 42.5°C